The answer is B. two chlorine atoms.
The chemical formula of the compound is BaCl2~
Answer:
Radon
Explanation:
One element that will have similar property as an atom with 118 protons is Radon because it will belong to the p-block and the noble gas group.
The element will follow atomic number 117 which is already in the 7th group on the periodic table.
- Generally, on the periodic table, elements in the same group will have the same chemical property.
- The valency of most elements determines their chemical behavior.
- Since our mystery element is in the 8th group, noble gas group, it will mostly behave like any of the elements in the group.
Answer:
a) The minimum number of residues in alfa helix to span 30 Å, is 20 residues
b) The minimum number of residues in a Beta sheet to span 30 Å, is 8.6 residues.
c) Should be better if you show an image to do this question.
d) Look answer below.
Explanation:
a) Due to every 5,4 Å there are 3,6 residues. So doing the calculus give us 20 residues.
b) That’s it because every 7 Å there are 2 residues
c) First of all, is important to know that the residues of a protein define the secondary structure due to intermolecular and intramolecular interactions. Taking in account that the membrane bilayer has a lipid interior α-helical domains which are hydrophobic principally are embedded in membranes by hydrophobic interactions with the lipid interior of the bilayer favored by Van der Waals interactions and probably also by ionic interactions with the polar head groups of the phospholipids.
The molarity of the HCl is 1 M when 12.0 of .500 M NaOH neutralized 6.0 ml of HCl solution.
Explanation:
Data given:
molarity of the base NaOH, Mbase =0. 5 M
volume of the base NaOH, Vbase = 12 ml
volume of the acid, Vacid = 6 ml
molarity of the acid, Macid = ?
The titration formula for acid and base is given as:
Mbase Vbase = Macid Vacid
Macid =
Macid = 1 M
we can see that 1 M solution of HCl was used to neutralize the basic solution of NaOH. The volume of NaOH is 12 ml and volume of HCl used is 6ml.
